Free GMAT Problem Solving Practice Question

This is a reflection, or mirror, problem. The robot must touch the line y = 2 once, and both P and Q lie above it (y-coordinates 4 and 9 are both greater than 2). To minimize a path that bounces off a line, reflect one endpoint across the line and draw a straight segment: any path from P to a point R on the line and then to Q has the same length as the path from the reflected point P′ to R to Q, since reflecting does not change the length of the first leg. The shortest such path is the straight line from P′ to Q, which automatically picks the best touch point.

Reflecting P across y = 2 keeps the x-coordinate and sends y to (2×2) − y, so P = (1, 4) becomes P′ = (1, 0). The straight-line distance from P′ = (1, 0) to Q = (13, 9) has horizontal change 12 and vertical change 9, so it is √(12² + 9²) = √(144 + 81) = √225 = 15. The segment from P′ to Q crosses y = 2 at an x between 1 and 13, so a valid touch point exists and 15 is actually reached. So the answer is C.

Among the traps: A (9) sums the two vertical drops to the line and forgets the points are far apart horizontally; B (13) is the straight P-to-Q distance from someone who never reflects and ignores that the path must reach the line; D (17) adds the horizontal gap to the direct vertical gap, a right-angle reading of the direct route; E (21) is the literal down-across-up path, which touches the line but is longer than the straight reflected route. The key idea is that the minimum uses the sum of the two heights above the line (2 + 7 = 9 as the vertical leg), not the difference.